Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Solve the differential equation.**
\[ y' + 7xe^y = 0 \]
To solve this differential equation, we will employ the method of separation of variables. The given equation is:
\[ y' + 7xe^y = 0 \]
First, rewrite \( y' \) as \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} + 7xe^y = 0 \]
Next, separate the variables \( y \) and \( x \):
\[ \frac{dy}{dx} = -7xe^y \]
\[ \frac{dy}{e^y} = -7x \, dx \]
Integrate both sides to get the solution:
\[ \int \frac{1}{e^y} \, dy = \int -7x \, dx \]
The left side integral with respect to \( y \) is:
\[ \int e^{-y} \, dy = -e^{-y} \]
The right side integral with respect to \( x \) is:
\[ \int -7x \, dx = -7 \cdot \frac{x^2}{2} = -\frac{7x^2}{2} \]
Equating both integrals, we have:
\[ -e^{-y} = -\frac{7x^2}{2} + C \]
where \( C \) is the constant of integration.
To simplify, we multiply through by \(-1\):
\[ e^{-y} = \frac{7x^2}{2} + C \]
This gives us the implicit solution to the differential equation. Depending on the problem's context, further manipulation or solving for \( y \) in terms of \( x \) might be required, but as it stands, this is the general form of the solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faea7a391-77a2-4be9-b3a3-98e0f70bf810%2F8781a417-fb8e-4993-81bd-622bbcc6bce4%2Fescbjcp_processed.png&w=3840&q=75)
Transcribed Image Text:**Solve the differential equation.**
\[ y' + 7xe^y = 0 \]
To solve this differential equation, we will employ the method of separation of variables. The given equation is:
\[ y' + 7xe^y = 0 \]
First, rewrite \( y' \) as \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} + 7xe^y = 0 \]
Next, separate the variables \( y \) and \( x \):
\[ \frac{dy}{dx} = -7xe^y \]
\[ \frac{dy}{e^y} = -7x \, dx \]
Integrate both sides to get the solution:
\[ \int \frac{1}{e^y} \, dy = \int -7x \, dx \]
The left side integral with respect to \( y \) is:
\[ \int e^{-y} \, dy = -e^{-y} \]
The right side integral with respect to \( x \) is:
\[ \int -7x \, dx = -7 \cdot \frac{x^2}{2} = -\frac{7x^2}{2} \]
Equating both integrals, we have:
\[ -e^{-y} = -\frac{7x^2}{2} + C \]
where \( C \) is the constant of integration.
To simplify, we multiply through by \(-1\):
\[ e^{-y} = \frac{7x^2}{2} + C \]
This gives us the implicit solution to the differential equation. Depending on the problem's context, further manipulation or solving for \( y \) in terms of \( x \) might be required, but as it stands, this is the general form of the solution.
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